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Let $p_n$ be the largest prime divisor of $n^4+n^2+1$ and let $q_n$ be the largest prime divisor of $n^2+n+1$. Then $p_n=q_{n^2}$, and from $$n^4+n^2+1=(n^2+1{)}^2-n^2=(n^2-n+1)(n^2+n+1)=((n-1{)}^2+(n-1)+1)(n^2+n+1)$$ it follows that $p_n=\max \{q_n,q_{n-1}\}$ for $n\ge 2$. Keeping in mind that $n^2-n+1$ is odd, we have $$\gcd (n^2+n+1,n^2-n+1)=\gcd (2n,n^2-n+1)=\gcd (n,n^2-n+1)=1.$$ Therefore, $q_n\ne q_{n-1}$. To prove the result, it suffices to show that the set $$S=\{n\in {\mathbb{Z}}_{\ge 2}\mid q_n>q_{n-1}\text{ and }{q}_n>q_{n+1}\}$$ is infinite, since for each $n\in S$ one has $$p_n=\max \{q_n,q_{n-1}\}=q_n=\max \{q_n,q_{n+1}\}=p_{n+1}.$$ Suppose on the contrary that $S$ is finite. Since $q_2=7<13=q_3$ and $q_3=13>7=q_4$, the set $S$ is non-empty. Since it is finite, we can consider its largest element, say $m$. Note that it is impossible that $q_m>q_{m+1}>q_{m+2}>\dots$ because all these numbers are positive integers, so there exists a $k\ge m$ such that $q_k<q_{k+1}$ (recall that $q_k\ne q_{k+1}$). Next observe that it is impossible to have $q_k<q_{k+1}<q_{k+2}<\dots$, because $$q_{(k+1{)}^2}=p_{k+1}=\max \{q_k,q_{k+1}\}=q_{k+1},$$ so let us take the smallest $l\ge k+1$ such that $q_l>q_{l+1}$. By the minimality of $l$ we have $q_{l-1}<q_l$, so $l\in S$. Since $l\ge k+1>k\ge m$, this contradicts the maximality of $m$, and hence $S$ is indeed infinite.